equations imply equation

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July 5th 2006, 06:27 PM
malaygoel

equations imply equation

Prove that the equations
$x^2 - 3xy + 2y^2 + x - y = 0$
and $x^2 - 2xy + y^2 - 5x + 7y = 0$
imply the equation
$xy - 12x + 5y = 0$ .

Please help.

Keep Smiling
Malay
July 6th 2006, 10:21 AM
Soroban

Hello, Malay!

I'm baffled . . . (more than usual)

Quote:

Prove that the equations:
$x^2 - 3xy + 2y^2 + x - y = 0$ and $x^2 - 2xy + y^2 - 5x + 7y = 0$
imply the equation: $xy - 12x + 5y = 0$

I thought this would be simple (perhaps imaginative) algebra,
. . but I've made absolutely no progress.

Then it occured to me that I have never done a problem like this.
Two equations imply another? . . . Do I understand what that means?
(I don't think I do . . . )

I can find the intersections of the two graphs, but that's not the mission.

The first equation represents a pair of intersecting lines.
The second is a parabola (tipped 45°).

Somehow these two imply the hyperbola: $y\:=\:\frac{12x}{x+5}$

How does that work? .I have no idea of what I'm doing . . .
July 6th 2006, 11:20 AM
ThePerfectHacker

Quote:

Originally Posted by Soroban

Then it occured to me that I have never done a problem like this.
Two equations imply another? . . . Do I understand what that means?
(I don't think I do . . . )

It means, that if $x,y$ are real numbers that satisfy those equations then (implies) they satisfy a new equation.

For example, if $x+2y=1$ and $2x+y=1$ implies that they satisfy, $3x+3y=2$
July 6th 2006, 01:14 PM
Soroban

Hello, TPHacker!

Thank you for the explanation.

You're saying: If $(x,y)$ satisfies the two given equations,
. . then $(x,y)$ satisfies $xy - 12x + 5y\,=\,0$

I suspected that that was an interpretation of "implies"
. . but it doesn't work.

I found three points that satisfy the two given equations:
. . $(0,0),\;(3,2),\;(5,3)$

But only $(0,0)$ satisfies: $xy - 12x + 5y\,=\,0$
July 6th 2006, 01:55 PM
ThePerfectHacker

Quote:

Originally Posted by Soroban

I found three points that satisfy the two given equations:
. . $(0,0),\;(3,2),\;(5,3)$

But only $(0,0)$ satisfies: $xy - 12x + 5y\,=\,0$

This explains why you had no way of demonstrating this, because it was wrong!
July 7th 2006, 01:43 AM
JakeD

Quote:

Originally Posted by Soroban

Then it occured to me that I have never done a problem like this.
Two equations imply another? . . . Do I understand what that means?
(I don't think I do . . . )

Quote:

Originally Posted by ThePerfectHacker

It means, that if $x,y$ are real numbers that satisfy those equations then (implies) they satisfy a new equation.

For example, if $x+2y=1$ and $2x+y=1$ implies that they satisfy, $3x+3y=2$

Generalizing TPH's reply, this problem can be seen one of vector spaces and linear algebra.

We are given three equations in the vector space V of equations of the form $a_1 x^2 + a_2 xy + a_3 y^2 + a_4 x + a_5 y = 0.$ V is a vector space because linear combinations of equations in V are again in V.

The three given equations

$x^2 - 3xy + 2y^2 + x - y = 0$
$x^2 - 2xy + y^2 - 5x + 7y = 0$
$xy - 12x + 5y = 0$

can be represented as coefficient vectors $[a_1, \ldots, a_5].$ Putting those 3 vectors into a matrix

$\bmatrix{ 1 & -3 & 2 &1 &-1 \\ 1 & 2 & 1 &-5 &7 \\ 0 & 1 & 0 &-12 &5\end{bmatrix}$

the problem becomes determining whether or not the third row (equation) is a linear combination of (implied by) the first two rows (equations). The answer is no because there is no way to take a non-trivial linear combination of the first and second rows and get zeroes in both the first and third columns as in the third row. Put more abstractly, the third equation is not in the subspace spanned by the first two equations, so the third equation is not implied by the first two equations.

Back at TPH's example, the third equation is a linear combination (the sum) of the first two equations, so the third equation is implied by the first two equations.

Well, this was supposed to make the answer more obvious, but it turned out to be a long-winded way to do that! At the least I hope it makes the problem a familiar one from linear algebra.
July 7th 2006, 06:33 AM
ThePerfectHacker

JakeD, I just luv your approach, but I do not understand what you are doing. Can you in steps explain what is going on. Begin by defining the vector space....
Also, does the fact
$ \bmatrix{1 & -3 & 2 &1 &-1 \\1 & 2 & 1 &-5 &7 \\0 & 1 & 0 &-12 &5\end{bmatrix} $
Should be,
$ \bmatrix{1 & -3 & 2 &1 &-1 \\1 & -2 & 1 &-5 &7 \\0 & 1 & 0 &-12 &5\end{bmatrix} $
Make a difference in your proof?
July 8th 2006, 12:43 AM
malaygoel

it may help:
$x^3 - 3xy + 2y^2 + x - y$ $=(x-y)(x-2y+1)$

Keep Smiling
Malay
July 8th 2006, 12:48 AM
malaygoel

Quote:

Originally Posted by malaygoel

Prove that the equations
$x^2 - 3xy + 2y^2 + x - y = 0$
and $x^2 - 2xy + y^2 - 5x + 7y = 0$
imply the equation
$xy - 12x + 5y = 0$ .

Please help.

Keep Smiling
Malay

there is one correction.
The coefficient of y in $xy - 12x + 5y = 0$ is 15 not 5.
I am sorry for it.

Keep Smiling
Malay
July 8th 2006, 04:06 AM
malaygoel

Let me try once.
We have two equations
$(x-y)(x-2y+1)=0$
$x^2 - 2xy + y^2 -5x + 7y = 0$
If x=y, we have the case x=y=0 which the third equation satisfies.
$Second Case$
$x-2y+1=0$
Multiplying both sides by $(2x-y)$
We have $2x^2 - 5xy + y^2 +2x -y=0$
This equation and the second equation $(x^2 - 2xy + y^2 -5x + 7y = 0)$
produces the third equation.
Is there any other method?

Keep Smiling
Malay
July 8th 2006, 05:30 AM
JakeD

Quote:

Originally Posted by ThePerfectHacker

JakeD, I just luv your approach, but I do not understand what you are doing. Can you in steps explain what is going on.

Unfortunately my approach was incorrect. The error was in these two statements.

Quote:

Originally Posted by JakeD

The problem becomes determining whether or not the third row (equation) is a linear combination of (implied by) the first two rows (equations)... Put more abstractly, the third equation is not in the subspace spanned by the first two equations, so the third equation is not implied by the first two equations.

That the third equation is not a linear combination of the first two equations does not determine that the third equation is not implied by them. I found this generally only works for linear equations. :o
July 8th 2006, 07:31 AM
Soroban

Hello, all!

I found another approach . . . but hit a wall.
. . Maybe someone can carry on . . .

We want: . $\begin{pmatrix}[1]\;x^2-3xy + 2y^2 + x - y \:=\:0\\ [2]\;x^2-2xy + y^2 - 5x + 7y \:= \:0\end{pmatrix}$ $\;\Longrightarrow \;[3]\; xy - 12x + 15y \:=\:0$

$[1]\;x^2 - 3xy + 2y^2 + x - y \:=\:0\quad\Rightarrow\quad x^2$ $- 2xy + y^2 - xy + y^2 + x - y \:= \:0$

. . $\Rightarrow\quad (x-y)^2 - y(x - y) + (x - y)\:=\:0\quad\Rightarrow\quad (x-y)^2$ $- (x-y)(y-1)\:=\:0$

. . $\Rightarrow\quad(x - y)^2 \:=\:(x-y)(y-1)$ [4]

$[2]\;x^2-2xy+y^2-5x+7y\:=\:0\quad\Rightarrow\quad (x-y)^2$ $\:=\:-(5x - 7y) \:=\:0$

. . $\Rightarrow\quad (x - y)^2\:=\:5x - 7y$ [5]

Equate [4] and [5]: . $(x - y)(y - 1) \:=\:5x - 7y\quad\Rightarrow\quad xy$ $-\,6x + 8y - y^2\:=\:0$

This is not $[3]$ , of course, but it's closer than anything I've found so far.

If only I could show that $y^2 \,= \,6x-7y$ , but no luck yet.

Anyone? .Anyone?
July 8th 2006, 08:09 AM
galactus

I will put in my 2 cents, for what it's worth. I doubt if it amounts to anything, but we'll see.

We have $x^{2}-3xy+2y^{2}+x-y=0$ ....[/1]

$x^{2}-2xy+y^{2}-5x+7y=0$ ......[2]

$xy-12x+15y=0$ ...........[3]

Solve [1] for x and we have $x=2y-1\,\ and\,\ x=y$

Sub $x=2y-1$ into [2] and simplify:

$y^{2}-5y+6=(y-3)(y-2)$ .....[4]

Sub into [3] and simplify:

$2y^{2}-10y+12=2(y-3)(y-2)$ .........[5]

Note that [5] is twice [4]

Solutions are (3,2) and (5,3).

(5,3) is the only case which produces 0 in all equations, though.

Well, except for (0,0).

Just a thought.
July 8th 2006, 11:06 AM
JakeD

Quote:

Originally Posted by galactus

I will put in my 2 cents, for what it's worth. I doubt if it amounts to anything, but we'll see.

We have $x^{2}-3xy+2y^{2}+x-y=0$ ....[/1]

$x^{2}-2xy+y^{2}-5x+7y=0$ ......[2]

$xy-12x+15y=0$ ...........[3]

Solve [1] for x and we have $x=2y-1\,\ and\,\ x=y$

Sub $x=2y-1$ into [2] and simplify:

$y^{2}-5y+6=(y-3)(y-2)$ .....[4]

Sub into [3] and simplify:

$2y^{2}-10y+12=2(y-3)(y-2)$ .........[5]

Note that [5] is twice [4]

Solutions are (3,2) and (5,3).

(5,3) is the only case which produces 0 in all equations, though.

Well, except for (0,0).

Just a thought.

I put the equations into a spreadsheet and found (3,2), (5,3) and (0,0) are all solutions to all the equations. So galactus it seems your algebra is correct but the arithmetic is not.

Soroban found these these same solutions before the error in the third equation was revealed.

Quote:

Originally Posted by Soroban

Hello, TPHacker!

Thank you for the explanation.

You're saying: If $(x,y)$ satisfies the two given equations,
. . then $(x,y)$ satisfies $xy - 12x + 5y\,=\,0$

I suspected that that was an interpretation of "implies"
. . but it doesn't work.

I found three points that satisfy the two given equations:
. . $(0,0),\;(3,2),\;(5,3)$

But only $(0,0)$ satisfies: $xy - 12x + 5y\,=\,0$
July 11th 2006, 04:35 AM
malaygoel

Quote:

Originally Posted by Soroban

Hello, all!

I found another approach . . . but hit a wall.
. . Maybe someone can carry on . . .

We want: . $\begin{pmatrix}[1]\;x^2-3xy + 2y^2 + x - y \:=\:0\\ [2]\;x^2-2xy + y^2 - 5x + 7y \:= \:0\end{pmatrix}$ $\;\Longrightarrow \;[3]\; xy - 12x + 15y \:=\:0$

$[1]\;x^2 - 3xy + 2y^2 + x - y \:=\:0\quad\Rightarrow\quad x^2$ $- 2xy + y^2 - xy + y^2 + x - y \:= \:0$

. . $\Rightarrow\quad (x-y)^2 - y(x - y) + (x - y)\:=\:0\quad\Rightarrow\quad (x-y)^2$ $- (x-y)(y-1)\:=\:0$

. . $\Rightarrow\quad(x - y)^2 \:=\:(x-y)(y-1)$ [4]

$[2]\;x^2-2xy+y^2-5x+7y\:=\:0\quad\Rightarrow\quad (x-y)^2$ $\:=\:-(5x - 7y) \:=\:0$

. . $\Rightarrow\quad (x - y)^2\:=\:5x - 7y$ [5]

Equate [4] and [5]: . $(x - y)(y - 1) \:=\:5x - 7y\quad\Rightarrow\quad xy$ $-\,6x + 8y - y^2\:=\:0$

This is not $[3]$ , of course, but it's closer than anything I've found so far.

If only I could show that $y^2 \,= \,6x-7y$ , but no luck yet.

Anyone? .Anyone?

Watch your final equation.
it is just the differnce of first two equations. But your work is not for wastebasket.
You can break the wall for the case x is not equal to y using your 4th and 5th equations.
If x is equal to y, then it is a trivial case since then x=y=0.

Keep Smiling
Malay